∫∠ACB = 90, D is the midpoint of AB,

∴DC=DA=DB,∴∠B=∠DCB.

And: ∫ ABC △ FDE, ∴ ∠ FDE = ∠ B.

∴∠FDE=∠DCB,∴DG∥BC.

∴∠AGD=∠ACB=90 ,∴DG⊥AC.

And ∵DC=DA, ∴G is the midpoint of AC,

∴cg= 12ac= 12×8=4,dg= 12bc= 12×6=3，

∴S△DGC= 12CG？ DG= 12×4×3=6。

(2) Cooperation and communication

Solution 1: as shown in the following figure:

∵△ABC≌△FDE,∴∠B=∠ 1.

∠∠c = 90 ,ed⊥ab，

∴∠A+∠B=90，∠A+∠2=90，

∴∠B=∠2,∴∠ 1=∠2，

∴GH=GD.

∠∠A+∠2 = 90，∠ 1+∠3=90，

∴∠A=∠3,∴AG=GD，

∴AG=GH, that is, point G is the midpoint of AH.

In Rt△ABC, AB=AC2+BC2=82+62= 10.

∫d is the midpoint of AB, ∴ ad = 12ab = 5.

In △ADH and △ACB, ∫∠A =∠A, ∠ ADH = ∠ ACB = 90,

∴△ADH∽△ACB, ∴ ADAC = DHCB, that is, 58 = DH6, and the solution is DH= 154.

∴s△dgh= 12s△adh= 12× 12×dh？ AD = 14× 154×5 = 75 16。

Solution 2: identical solution 1, and g is the midpoint of AH.

Connecting BH, ∵DE⊥AB, D is the midpoint of AB,

∴。 Let AH=x, then ch = 8-X.

In Rt△BCH, CH2+BC2=BH2.

Namely: (8-x)2+36=x2.